Difference between revisions of "2009 AIME I Problems/Problem 1"
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For the maximum, we have that <math>b^2 = 9c</math>; <math>b</math> is maximized when <math>9c</math> is the greatest possible perfect square; this happens when <math>c = 4</math>, yielding <math>b = 6</math>. Thus, the largest possible geometric number is 964. | For the maximum, we have that <math>b^2 = 9c</math>; <math>b</math> is maximized when <math>9c</math> is the greatest possible perfect square; this happens when <math>c = 4</math>, yielding <math>b = 6</math>. Thus, the largest possible geometric number is 964. | ||
− | Our answer is thus <math>964 - 124 = | + | Our answer is thus <math>964 - 124 = 840. |
==Solution 3== | ==Solution 3== | ||
− | The smallest geometric number is 124 because 123 and any number containing a zero does not work. 964 is the largest geometric number because the middle digit cannot be 8 or 7. Subtracting the numbers gives <math>\boxed{840}. | + | The smallest geometric number is 124 because 123 and any number containing a zero does not work. 964 is the largest geometric number because the middle digit cannot be 8 or 7. Subtracting the numbers gives </math>\boxed{840}.$ |
==Video Solution== | ==Video Solution== |
Revision as of 20:10, 11 May 2020
Problem
Call a -digit number geometric if it has distinct digits which, when read from left to right, form a geometric sequence. Find the difference between the largest and smallest geometric numbers.
Solution 1
Assume that the largest geometric number starts with a nine. We know that the common ratio must be a rational of the form for some integer , because a whole number should be attained for the 3rd term as well. When , the number is . When , the number is . When , we get , but the integers must be distinct. By the same logic, the smallest geometric number is . The largest geometric number is and the smallest is . Thus the difference is .
Solution 2
Consider the three-digit number . If its digits form a geometric sequence, we must have that , that is, .
The minimum and maximum geometric numbers occur when is minimized and maximized, respectively. The minimum occurs when ; letting and achieves this, so the smallest possible geometric number is 124.
For the maximum, we have that ; is maximized when is the greatest possible perfect square; this happens when , yielding . Thus, the largest possible geometric number is 964.
Our answer is thus $964 - 124 = 840.
==Solution 3== The smallest geometric number is 124 because 123 and any number containing a zero does not work. 964 is the largest geometric number because the middle digit cannot be 8 or 7. Subtracting the numbers gives$ (Error compiling LaTeX. Unknown error_msg)\boxed{840}.$
Video Solution
~IceMatrix
See also
2009 AIME I (Problems • Answer Key • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.